At first I would like to apologize to the members who answered this. This question had considerable information so I asked different question.
Actual question I wanted to ask,
We know that $$\int (x + b)^n\,\mathrm dx =\frac{(x+b)^{n+1}}{n+1} + C$$
so while solving $$\int x^2\,\mathrm dx = \frac13x^3 + C$$
also, $$\int (x + 3)^4\,\mathrm dx = \frac15(x + 3)^5 + C$$
but for trig functions like, $$\int \sin^2x\,\mathrm dx \neq \frac13\sin^3x + C$$
Instead, we have to simplify it first algebraically. So why does the intuition fail here?
I apologize once again. But still I learnt some new things.
The 'nice' form you see from the polynomial integration is a result of a rigorous derivation, which by right isn't supposed to become something 'intuitive'...
Also actually the trigonometric functions have their own polynomial 'equivalents' so to say.
For example, $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots$$
This probably would explain why $$\int \sin^2x\,\mathrm dx \neq \frac13\sin^3x + C$$ since $$\int \left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots\right)^2\,\mathrm dx \neq \frac13\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots\right)^3 + C$$